Let Rbe a simply connected region in the xy-plane. Show that the portion of the paraboloid with equation $z=x^2+y^2$ determined by R has the same area as the portion of the saddle with equation $z=x^2-y^2$determined by R.

Let givenRbe a simply connected region in the xy-plane.

${\int }_{s}1dS=\int {\int }_R\sqrt{{\left(\frac{\partial z}{\partial x}\right)}^2+{\left(\frac{\partial z}{\partial y}\right)}^2+1dA}...........\left(1\right)$

$$\begin{gathered} \frac{\partial z}{\partial x}=\frac{\partial }{\partial x}(x^2+y^2) \\ =2x \end{gathered}$$

and,

$$\begin{gathered} \frac{\partial z}{\partial y}=\frac{\partial }{\partial y}(x^2+y^2) \\ =2y \end{gathered}$$

Then,

$$\begin{gathered} dS=\left(\sqrt{{\left(\frac{\partial z}{\partial x}\right)}^2+{\left(\frac{\partial z}{\partial y}\right)}^2+1}\right)dA \\ =\left(\sqrt{{\left(2x\right)}^2+{\left(2y\right)}^2+1}\right)dA \\ =\left(\sqrt{4x^2+4y^2+1}\right)dA \end{gathered}$$

Then,by(1),the surface area of the paraboloid surface I will be,

$$\begin{gathered} {\int }_{s}1dS=\int {\int }_R\sqrt{{\left(\frac{\partial z}{\partial x}\right)}^2+{\left(\frac{\partial z}{\partial y}\right)}^2+1dA} \\ =\int {\int }_R\left(\sqrt{4x^2+4y^2+1}\right)dA \end{gathered}$$

Here, the surface S is the portion of the following paraboloid; $z=x^2-y^2$

Now, first find $\frac{\partial z}{\partial x}$and $\frac{\partial z}{\partial x}$. The first partial derivatives of z are,

$$\begin{gathered} \frac{\partial z}{\partial x}=\frac{\partial }{\partial x}(x^2-y^2) \\ =2x \end{gathered}$$

And,

$$\begin{gathered} \frac{\partial z}{\partial y}=\frac{\partial }{\partial y}(x^2-y^2) \\ =-2y \end{gathered}$$

Then,

$$\begin{gathered} dS=\left(\sqrt{{\left(\frac{\partial z}{\partial x}\right)}^2+{\left(\frac{\partial z}{\partial y}\right)}^2}+1\right)dA \\ =\left(\sqrt{{\left(2x\right)}^2+{\left(-2y\right)}^2+1}\right)dA \\ =\left(\sqrt{4x^2+4y^2+1}\right)dA \end{gathered}$$

Then,by(1),the surface area of the paraboloid surface II will be,

$$\begin{gathered} {\int }_{s}1dS=\int {\int }_R\sqrt{{\left(\frac{\partial z}{\partial x}\right)}^2+{\left(\frac{\partial z}{\partial y}\right)}^2+1dA} \\ =\int {\int }_R\left(\sqrt{4x^2+4y^2+}\right)dA \end{gathered}$$

For each surface,the surface area is,

${\int }_{s}1dS=\int {\int }_R\left(\sqrt{4x^2+4y^2+1}\right)dA$

Where,the limits determined by R,

Thus,it is proved that the areas of both surface are equal

Hence proved